TITLE: Topological classification of a finite union of open balls in $\mathbb{R}^n$
QUESTION [6 upvotes]: What are the possible topological shapes (i.e. up to homeomorphism) of a finite union of open balls in $\mathbb{R}^n?$
For example, for $n = 1$, open balls are just open intervals and a finite union of open intervals is just a disjoint union of open intervals (the union of two intervals that overlap is again an interval)
For $n = 2$, we can have a disjoint union of open balls, but also an annulus, or an "annulus with many holes", a disjoint union of "annuli with many holes" and maybe something else. 
Is there a classification of these possible shapes? 
EDIT

REPLY [3 votes]: First a definition: A manifold $M$ is called tame if it is homeomorphic to the interior of a compact manifold $\bar{M}$ with boundary. 
A connected tame surface $M$ is characterized by its genus (i.e. genus of $\bar{M}$), orientability and the number of boundary components of $\bar{M}$. 
For $n=2$ the classification will be: A surface $S$ is homeomorphic to a finite union of open round disks in $R^2$ if and only if $S$ is a tame surface of genus zero. 
In higher dimensions things are more complicated but one can prove that an $n$-dimensional manifold $M$ is homeomorphic to a finite union of open balls in $R^n$ if and only if $M$ is tame and the manifold $\bar{M}$ is homeomorphic to a piecewise-linear submanifold in $R^n$. 
In order to appreciate the complexity of this class of manifolds, note that every finite $k$-dimensional simplicial complex is homotopy-equivalent to a finite union of open round balls in $R^n$, $n=2k+1$. Already for $n=3$, a topological classification of finite unions  of round balls in $R^3$ is hopeless.